I'm trying to make Cu(NO3)2. I made a small amount using Cu metal and Ag(NO3)2. But this of course would not be an economical way of making it.
I have some Ca(NO3)2, so I figured with CuSO4 this would precipitate CaSO4 and leave Cu(NO3)2. So I mixed up approximately 1M solutions of both
CuSO4 and Ca(NO3)2. When I mixed the two solutions together, I expected to get a thick precipitate (which is why I only used a concentration of about
1M). Instead, nothing appeared to happen! I was about to throw this solution away, but after leaving it sit for nearly an hour finally I saw a small
amount of precipitate gradually starting to form. So why is this reaction so darned slow? If I add dilute H2SO4 to a Ca(NO3) solution, a thick
precipitate forms immediately.
Another question has to do with the reaction between KI and CuSO4. I read that this is a way to prepare "highly pure" I. So I tried it and
sure enought I got some I - though not as quickly as I do using H2O2 in acidic solution. I also got an insoluable, chalky precipitate. What would
this be? Is it CuSO3 maybe? Why is CuSO4 recommended for preparing I vs. using H2O2?vulture - 23-2-2004 at 00:29
CaSO4 is not that insoluble, it could very well be soluble enough in your diluted solution. What's diluted H2SO4? 1M? 5M?
2KI + CuSO4 ---> CuI2 + K2SO4
The precipitate should be CuI2.I am a fish - 23-2-2004 at 04:02
Quote:
Originally posted by vulture
2KI + CuSO4 ---> CuI2 + K2SO4
The precipitate should be CuI2.
Wrong!
4KI + 2CuSO4 --> 2CuI + I2 + 2K2SO4
The precipitate is copper(I) iodide.
[Edited on 23-2-2004 by I am a fish]vulture - 23-2-2004 at 04:51
Interesting, I've never seen this reaction before.
Rather uneconomical to make I2 this way, because you "loose" 50%.Marvin - 23-2-2004 at 13:36
Incidentally, its a quantitive reaction so its very useful analytically. You add excess iodide and then titrate with starch/thiosulphate to determine
amounts of copper(II).hodges - 23-2-2004 at 16:20
Quote:
Originally posted by I am a fish
4KI + 2CuSO4 --> 2CuI + I2 + 2K2SO4
The precipitate is copper(I) iodide.
Wow, I would have never guessed. I figured that since I was being oxidized, something had to be reduced, that's why I guessed SO4 might be
reduced. Never stopped to think about the Cu being reduced.
I don't know why I read in several different places that CuSO4/KI is a good way to prepare iodine. Given only 50% efficiency. I was thinking
maybe CuI decomposes to I upon heating, but not so according to CRC handbook. Sure seems H2O2 + H+ + KI is better. I once used this reaction to
produce some I for an energetic material which also contains nitrogen . Anyway,
only thing I can think of is maybe the people recommending the CuSO4/KI only knew of that one reaction and not the other, or figured that CuSO4 would
be easier to get than H2O2/acid.
As far as the CuSO4/Ca(NO3)2 reaction. I tried it again with more concentrated solutions. The reaction was faster, but far from instantaneous. It
was about 15 seconds before any precipitate formed and about 15 minutes before it was complete. Anyway, I filtered out the CaSO4. I now have a
(presumably) Cu(NO3)2 solution. My solution looks slightly different from CuSO4. Not sure how to describe it - maybe not quite as deep of a blue.
Cu(NO3)2 may have a slightly different color than CuSO4 in solution, or it may just be impurities such as unfiltered CaSO4. In a couple weeks though
the water should evaporate and I can heat to see if I really have Cu(NO3)2. Heating Cu(NO3)2 prepared from Ag(NO3)2 and Cu resulted in brown NO2
being produced. Incidentally, heating Ca(NO3)2 alone did result in some NO2, but only a trace. So while Ca(NO3)2 may decompose to nitrogen oxides
upon heating as per its MSDS, the reaction does not seem to occur to anywhere near the extent of the reaction when heating Cu(NO3)2. And when I was
done heating the Cu(NO3)2, I had black copper oxide as expected. Tested in several ways - insoluable in water, made a blue CuSO4 solution with H2SO4
and a green CuCl2 solution with HCl.rikkitikkitavi - 28-2-2004 at 13:55
solubility of CaSO4 is abou 1gr/liter, so about 0,01 M . Thus 99 % would preciptate from a 1M solution. It could be that if the solutions are very
pure from the start, this allows for a supersaturation of the solution.
(about 100 times)
Increasing the concentration increases supersaturation and forces a quicker nucleation=> giving a preciptate.
The fact that the CaSO4 preciptates as a hydrated salt complicates things a bit to.
Try with 1 M solution, add a little solid CaSO4 and see if the preciptates formes faster. (the solid serves as a nucleation source)
Cu (NO3)2 decomposes at much lower temperature(about 200C) than Ca(NO3)2 (about 500 C). Acutally, sinc Cu(NO3)2 is a hydrated salt, the gases are NOx
and HNO3.smitty - 28-11-2004 at 09:56
apparently, the above reaction of KI with CuSO4 to yield I2, ie:
4KI+2CuSO4=>2CuI(precip)+I2(aq)+2K2SO4(aq)
is used because the resultant iodine extracted is very pure..
A question: how does one extract the I2 from the K2SO4 out of soln?neutrino - 28-11-2004 at 16:30
Filter out the CuI/I<sub>2</sub>, dry, and sublime the iodine out, condensing it on a cold surface.darkflame89 - 28-11-2004 at 18:50
Incidentally, I am a fish, you are wrong as well about the equation.(Edit: Ok, i didn't mean to be mean or anything, sorry) From my school
practical book, the correct formula for the formed copper(I) iodide is Cu2I2. I know its strange, I do not know why is it like that...
[Edited on 29-11-2004 by darkflame89]S.C. Wack - 28-11-2004 at 19:44
Quote:
Incidentally, I am a fish, you are wrong as well about the equation.
That is some strong language, especially considering that the Merck Index, the Handbook of Preparative Inorganic Chemistry, Ullmann's, Cotton and
Wilkinson's Advanced Inorganic Chemistry, Inorganic Syntheses, the CRC Handbook, Vogel's Handbook, Lange's Handbook, the Fiesers'
Reagents..., etc. give CuI as the formula for cuprous iodide.
Both formulas are well established and it doesn't matter who is "right". And no I'm not biased just because I refuse to call P2O5
anything else.smitty - 29-11-2004 at 09:09
thanks neutrino- though I was sure that the I2 was I2(aqueous) and the CuI(precipitate) in the equation, ie:
4KI+2CuSO4=>2CuI(precipitate)+I2(aqueous)+2K2SO4(aqueous).
could you perhaps give me more detail on the sublimation. Is Rikkitikkitavi on the right track with the saturated soln trick on the K2SO4 to
precipitate it out of the solution(despite confusing me a little with the (misprint?)CaSO4..I'm not a chemist..I did physics..only did 1st year
inorganic chem)smitty - 29-11-2004 at 09:18
whoops, sorry rikkitikkitavi..that's me not reading back on the thread properly. In my last message..delete everything after
"sublimation"..neutrino - 29-11-2004 at 14:36
Sublimation is simple: a solid is heated at point A and sublimes (passes directly from a solid to a gas). The vapor travels to a cold point B where it
is deposited (gas ->solid). For practical details, just search around. It is a very common method.
. Anyway,
only thing I can think of is maybe the people recommending the CuSO4/KI only knew of that one reaction and not the other, or figured that CuSO4 would
be easier to get than H2O2/acid.